Question

A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the solution

Answer 1

c = 1.14 mol/L; b = 1.03 mol/kg

Molar concentration

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

c = 1.14 mol/1 L = 1.14 mol/L

Molal concentration

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

b = 1.14 mol/1.106 kg = 1.03 mol/kg

Answer 2

Answer : The molarity and molality of the solution is, 1.00 mole/L and 0.904 mole/Kg respectively.

Solution : Given,

Density of solution = 1.19 g/ml

Molar mass of sodium carbonate (solute) = 84 g/mole

7.06% aqueous solution of sodium bicarbonate means that 7.06 gram of sodium bicarbonate is present in 100 g of solution.

Mass of sodium bicarbonate (solute) = 7.06 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 7.06 = 92.94 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.19g/ml}=84ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Moles of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{7.06g\times 1000}{84g/mole\times 84ml}=1.00mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Moles of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{7.06g\times 1000}{84g/mole\times 92.94g}=0.904mole/Kg$

Therefore, the molarity and molality of the solution is, 1.00 mole/L and 0.904 mole/Kg respectively.